# Mass of a photon

1. Jul 26, 2014

### prashant.d000

I have question on my own understanding.. So please correct me..

1. It is said that mass of photon is zero; Or it is so small that it can be neglected without any problems.
2. An object cannot surpass speed of light, because as it approaches to light speed, it's mass increases, and it require much more energy to accelerate it more. At light speed, the mass of object in Infinite!

My question is, Why the mass of photon is not infinite at light speed?

2. Jul 26, 2014

### ChrisVer

Although 2 is not quiet a good statement, because it deals with the relativistic mass and not the rest mass, the answer is simpler: because the photon travels at speed of light, it has to have zero rest mass- it's not accelerated to c. So far the photon's mass is absolutely zero, and not "negligible" as for example the neutrinos.

3. Jul 26, 2014

### Chronos

The photon has zero rest mass. All massless particles travel at c.

4. Jul 26, 2014

### Staff: Mentor

There's a FAQ at the top of this forum: https://www.physicsforums.com/showthread.php?t=511175 [Broken]

Last edited by a moderator: May 6, 2017
5. Jul 31, 2014

### Imager

I thought the concept of rest mass wasn’t applicable to a photon.

6. Jul 31, 2014

### Staff: Mentor

It's applicable, but it's better to call it "invariant mass" because that avoids the misconception that a particle must be capable of being at rest to have a "rest mass". What we're really talking about is the invariant length of the 4-momentum vector associated with the particle; photons have a perfectly well-defined 4-momentum just like other particles, and its length is zero, hence zero invariant mass.

7. Jul 31, 2014

### Imager

Did I also misunderstand that photons do not have inertial rest frames?

8. Jul 31, 2014

### Staff: Mentor

No, you understood that correctly. There is no inertial reference frame in which a photon is at rest.

In any inertial reference frame you can (in principle) measure the photon's energy and momentum, and calculate the quantity $mc^2 = \sqrt{E^2 - (pc)^2}$. The result is always zero, even though the values of E and p depend on the reference frame.

9. Jul 31, 2014

### Staff: Mentor

No, as jtbell said, you are correct that photons do not have inertial rest frames. It's just that they don't have to have them in order to have an invariant mass (which for photons is zero).

10. Jul 31, 2014

### Imager

Thank you PeterDonis and JTBell!

I think my problem was thinking backwards: photons do not have inertial reference frames, therefore they don’t have frame with invariant mass. Or maybe my problem is just thinking…

The statements:

“There is no inertial reference frame in which a photon is at rest,” ​

“What we're really talking about is the invariant length of the 4-momentum vector associated with the particle; photons have a perfectly well-defined 4-momentum just like other particles, and its length is zero, hence zero invariant mass.”​

make much more sense!

11. Aug 4, 2014

### SpiderET

It seems to me, that zero invariant mass (or rest mass) of photon is just an ad hoc setting of GR to make it work but it has nasty side effect of getting infinities or singularities many times in calculations.

Actually getting infinities in calculation of real objects is commonly a sign of bad theory, but in case of GR and getting infinite density of black hole makes GR not looking wrong for most people, but rather epic :)

What I know, it was never EXPERIMENTALLY proven that photon has a zero invariant mass and its only a pure theory. That doesnt mean, that GR havent been partially proven, but that is a completely different topic.

12. Aug 4, 2014

### Staff: Mentor

I'm not sure that's correct. I don't think GR cares whether light is massless or not. I think it only states that a massless object travels at c.

It doesn't make it a bad theory, but it does tell us that it is most likely incomplete.

We can't prove that photons are massless, we can only measure them to be under a certain mass based on the limits of our measurement methods and devices. Also, scientific theories can't be proven, as proofs only exist in math. They can only be verified to be accurate to some degree.

13. Aug 4, 2014

### Staff: Mentor

This is not correct. GR would work just fine with a massive photon. All that would be necessary would be to update the usual textbooks to use the term "invariant speed" rather than "speed of light" in some contexts.

Because any experimental measurement has some associated error it is logically impossible to prove equality through experiment. Regardless of the accuracy of the experiment, there would still be some range within which the measurement could turn out the same but the value would be not equal to the hypothesized value. This is not unique to the zero mass of a photon, but applies for any experimental measurement of any quantity with any theory where the theory predicts a definite value.

However, given that, there are incredibly sophisticated experiments which have measured the mass, and they have all found that it is 0 to within their experimental accuracy. I believe that the current upper limit on the mass of the photon is 6E-19 eV/c^2.

14. Aug 4, 2014

### SpiderET

Yes, thanks for pointing it out, currently there is an upper limit of mass of photon based on Coulomb law, but the other question is, if the theoretical prediction based on Proca equotations is really true.

15. Aug 5, 2014

### ChrisVer

Upper limit means upper limit. Apart from the fact that experiments will never give 0 as a result for the photon mass, the upper limit does also verify the 0-mass since it contains it.
which theoretical prediction of proca eqs?
The Proca Lagrangian is dealing with a massive spin-1 particle. It can be pretty good for Z or Ws particles. It can't be true for a photon except for taking the massless limit (at the correct point). If the photon was actually massive, it would also have 3 polarizations (just like the other bosons) which is actually not the case (the photon doesn't have a physical longitudial polarization).
There can be certain circumstances in which the photon can get a mass (eg in superconducting matterial) but in this case there's no reason for the Max.Eqs to be linear.

16. Aug 6, 2014

### KatamariDamacy

http://en.wikipedia.org/wiki/Photon#Experimental_checks_on_photon_mass

From this article it doesn't seem to me non-zero photon mass is completely unexpected or considered impossible, or that it would even change anything in our current equations. Kind of no big deal, or so it would seem.

17. Aug 6, 2014

### ChrisVer

And what does this article tell you? Apart from giving upper limits (extremely small values compared to all the other particles) to the photon mass...
It's pretty much massless.

18. Aug 6, 2014

### KatamariDamacy

It tells me that is an experimental question, that relativity would be unaffected, and that c would be a constant of nature for any object, rather than actual speed at which light moves.

It's not clear what is that actually supposed to mean or how it would work out, but it seems to me it means non-zero photon mass is not considered impossible and that it would not even change anything in our current equations. Do you know of something that would change if it turns out photons have some small but non-zero mass?

Last edited: Aug 6, 2014
19. Aug 6, 2014

### ChrisVer

and again I am repeating that zero mass is also verified by the upper bounds. The experiment will never give 0 as a result, because experiments don't give certain values but everything has an error width.
It is not going to change SR but people should actually break the U(1) gauge group of Electromagnetism somehow (by introducing a new higgs scalar). Also another thing is that c will just be a constant instead of a physical quantity (well that's an aesthetic thing).

20. Aug 6, 2014

### Staff: Mentor

I agree, but would go even further. It is not just pretty much, it is massless to within experimental accuracy. The upper bound describes the accuracy of the experiments, it does not describe the mass of the photon.

21. Aug 6, 2014

### Staff: Mentor

QED would be non-renormalizable.

22. Aug 6, 2014

### vanhees71

No, QED would stay renormalizable. In Abelian gauge theories you can in fact add a finite mass term for the gauge field by hand without spoiling gauge invariance and thus the proof of renormalizability goes through as for QED. The reason is the Stückelberg formalism for massive vector fields. I've explained this quickly recently in the following posting in the quantum section:

https://www.physicsforums.com/showpost.php?p=4813226&postcount=19

23. Aug 6, 2014

### Staff: Mentor

Interesting, thanks. I always thought that was the big theoretical problem.

24. Aug 6, 2014

### ChrisVer

25. Aug 8, 2014

### jlefevre76

Yeah, bear in mind, as quoted above, that while all experiments have some degree of experimental error, and we can only say it is zero to within the resolution of our instruments, all evidence up to this point indicates that photons do not have a mass.

And, rather than using that as a starting point for developing new theories, I'm willing to bet we'll need the next paradigm shift before we can correct theories like relativity or quantum physics. In the past, paradigm shifts that changed our understanding of physics have typically required a new form of math to describe observed behavior in a way no one had thought of before. For instance, calculus, linear algebra and differential equations were required to create Newton's law of gravitation, relativity, and quantum physics. I'm guessing (though no one can predict for sure) that the next revolution on physics will come not from finding an error in carefully taken measurements, but from applying a new form of math to unsolved problems.

That said, I've actually looked at the possibility of photons with mass. As already mentioned, since their relativistic mass is so much higher than their invariant mass could be at this point, it seems like that is not going to ever affect the outcome of future physics developments (as already mentioned, it doesn't affect any of our current equations). Am I wrong?